Inequations

ဒီအခန္းမွာေတာ့ Quadratic Inequations ေတြရဲ့ ေပးထားေသာ အေျခအေနကို ေျပလည္ေစမယ့္ solution set ကိုရွာမွာ ျဖစ္ပါတယ္။ Quadratic Inequations ေတြကို မေျဖရွင္းမီ Quadratic Function ဆိုတာကို သိရပါမယ္။

Quadratic Function

An expression f(x) = ax² + bx + c, where a, b, and c are numbers with a≠0 is

called a quadratic function.

ကိန္းရွင္တစ္ခုရဲ့ ႏွစ္ထပ္ကိန္း ပါ၀င္ေသာ function တစ္ခုကို quadratic function လို႔ေခၚပါတယ္။ Quadratic Function ရဲ့ characteristic က x² ပါ။ x² မပါရင္ quadratic function လို႔ မေျပာႏိုင္ပါဘူး။ ဒါေၾကာင့္ coefficient of x² ဟာ 0 မျဖစ္ရပါဘူး။

Quadratic Equation

ax² + bx + c = 0 is called a quadratic equation.

Solutions or Roots of Quadratic Equation





where a≠0









Graph of a Quadratic Function

The graph of a quadratic function is called a parabola. It is basically a curved shape opening up or down.



Quadratic Function တစ္ခုရဲ့ graph ကို parabola လို႔ေခၚပါတယ္။



When you have a quadratic function in the form f(x) = ax² + bx + c

if a > 0, then the parabola opens up ,

coefficient of x² ဟာ positive(a>0) ျဖစ္မယ္ဆိုရင္ graph ဆြဲတဲ့ အခါ open upward parabola ကိုရပါတယ္။

+x² => open upward parabola

if a <0, then the parabola opens down

coefficient of x² ဟာ positive(a<0) ျဖစ္မယ္ဆိုရင္ graph ဆြဲတဲ့ အခါ open downward parabola ကိုရပါတယ္။

- x² =>open downward parabola

Quadratic Inequations

The open sentences ax² + bx + c>0 and ax² + bx + c<0,a ≠ 0 are quadratic inequations in x.

The solution set of the quadratic inequations in x can be found by

(i) Algebraic method

(ii) Graphical method

The Binomial Theorem

Observe the following expansion.



(1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵



This expression can also be expressed as follows:













In general,if n is a positive integer,





















This expression is known as the Binomial Theorem.



Where the general coefficient or the coefficient of (r + 1)^th term is denoted as follows:

Note:

For the expansion of (x + y)ⁿ,where n is a positive integer,



(i) the binomial coefficients are all integers.

(ii) ⁿC₀ =ⁿC_n =1, ⁿC_n =ⁿC_n-1 = n, ⁿC_r =ⁿC_n-r .

The general term or the (r + 1)^th term of the expansion of (x + y)ⁿ is determined as follows:

(r + 1)^th term = ⁿC_r x^n-r y^r

Example 1

If the coefficients of x⁴ and x⁵ in the expansion of (3 + kx)¹⁰ are equal, find the value k.

(3 + kx)¹⁰ ကိုျဖန္႔္လွ်င္ x⁴ ၏ေျမွွာက္ေဖၚကိန္းသည္ x⁵ ၏ေျမွွာက္ေဖၚကိန္းႏွင့္ တူညီလွ်င္ k ၏တန္ဖိုးကို ရွာပါ။



Solution

(3 + kx)¹⁰

(r + 1)^th term = ¹⁰C_r 3^10-r (kx)^r = ¹⁰C_r 3^10-r k^rx^r



Here ¹⁰C_r 3^10-r k^r is the coefficient of x^r.

coefficient of x⁴ = ¹⁰C₄ 3⁶ k⁴

coefficient ofx⁵ = ¹⁰C₅ 3⁵ k⁵



By the problem,

¹⁰C₄ 3⁶ k⁴ = ¹⁰C₅ 3⁵ k<sup>4

10</sup>C₄ 3⁶ k⁴ = ¹⁰C₄ (6/5) 3⁵ k<sup>5

3 = (6/5)k

k = 5/2</sup>

Example 2

In the expansion of ,a x⁷ is four times the coefficient of x¹⁰. Find the value a.

ကိုျဖန္႔္လွ်င္ x⁷ ၏ေျမွွာက္ေဖၚကိန္းသည္ x¹⁰ ၏ေျမွွာက္ေဖၚကိန္း၏ 4 ဆျဖစ္လွ်င္ a ၏တန္ဖိုးကို ရွာပါ။



Solution







(r + 1)^th term =⁸C_r a^rx^16-3r

To get the coefficient of x⁷, 16-3r = 7 and r =3

To get the coefficient of x¹⁰, 16-3r = 10 and r =2



Therefore,coefficient of x⁷ = ⁸C₃ a³

coefficient of x¹⁰ = ⁸C₂ a²



By the problem,

⁸C₃ a³ = 4 ⁸C₂ a<sup>2

8</sup>C₂ (6/3) a³ = 4 ⁸C₂ a²

a = 2

Binomial Expansion

Binomial

A binomial is an expression that contains two terms.

Example: (a + b), (1 - a),k - h),(1 + x)²,(3x + 1)⁴.

The expansion of (1 + x)³ can be easily found:

(1 + x)³
= (1 + x)²(1 + x)

= (1 + 2x + x²) (1 + x)

= 1 + 2x + x² + x + 2x² + x³

= 1 + 3x + 3x² + x³

Similarly, we can use this method to find (1 + x)⁴, (1 + x)⁵ and (1 + x)⁶.


Look at the following expansions of (1 + x)ⁿ:

(1 + x)⁰ = 1

(1 + x)¹ = 1 + x

(1 + x)² = 1 + 2x + x²

(1 + x)³ = 1 + 3x + 3x² + x³

(1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴

(1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵

List the coefficients in a triangular array:

This array is known as the Pascal's Triangle.

Miscellaneous Problems

Find the remainder when 3x³ - x² + 7x + 5 is divided by 3x + 2.

Given that 2x³ - x² - 2x + 3 = (Ax + B)(x - 1)(x + 2) + C(x - 1) + D, find the values of A, B, C and D. Hence or otherwise, deduce the remainder when

2x³ - x² - 2x + 3 is divided by x² + x - 2.

Solve the equation 2x³ + x² - 19x = 6, giving your answers to two decimal places where necessary.

The expression x³ + px² + qx + 6 has the same remainder when divided by

x + 1 and 2 - x. Given that the remainder when the expression is divided by x + 3 is -60, find the value of p and of q.

By remainder theorem,

remainder = f(-2/3)

= 3(-2/3)³ - (-2/3)² + 7(-2/3) + 5

= -1

Let x = 1

2 - 1 - 2 + 3 = D

D = 2

Let x = -2

-16 - 4 + 4 + 3 = -3C + 2

3C = 15

C = 5

Let x = 0

3 = B (-1)(2) -5 + 2

-2B = 6

B = -3

Sub any value other than 1, 2 & 0 into x

A = 2

2x³ - x² - 2x + 3 ≡ (Ax + B)(x - 1)(x + 2) + C(x - 1) + D

Since x² + x - 2 = (x - 1)(x + 2), (note the similarity?)

thus, remainder = C(x - 1) + D

= 5(x - 1) + 2

= 5x - 3

2x³ + x² - 19x = 6

let f(x) = 2x³ + x² - 19x - 6

(x - 3) f(3) = 0

(x - 3) is a factor of f(x)

let f(x) = (x - 3)(Ax² + Bx + C)

A = 2

C = 2

let x = 1

2 + 1 - 19 - 6 = -2(2 + B + 2)

-22 = -8 - 2B

B = 7

f(x) = (x - 3)(2x² + 7x + 2)

2x² + 7x + 2 = 0

x =

= -3.19 or -0.31 (3 s.f.)

x = -3.19 or -0.31 or 3

let f(x) = x³ + px² + qx + 6

by remainder theorem, f(-1) = f(2)

-1 + p - q + 6 = 8 + 4p + 2q + 6

p - q - 5 = 4p + 2q + 14

3p + 3q = 9 --------(1)

by remainder theorem, f(-3) = -60

-27 + 9p - 3q + 6 = -60

3q = 39 + 9p --------(2)

sub (2) into (1):

3p + 39 + 9p = -9

12p = -48

p = -4

q = 1

Factor Theorem အသံုးခ် ပုစာၦမ်ား

Example 2

Given that x² + x - 6 is a factor of 2x⁴ + x³ - ax + bx + a + b - 1, find the value of a and b.

x² + x - 6 = (x + 3)(x + 2)

Let f(x) = 2x⁴ + x³ - ax² + bx + a + b - 1

f(-3) = 2(-3)⁴ + (-3)³ - a(-3)² - 3b + a + b - 1 = 0

134 - 8a - 2b = 0

4a + b = 67 --------(1)

f(2) = 2(2)⁴ + 2³ - a(2)² + 2b + a + b - 1 = 0

39 - 3a + 3b = 0

a - b = 13 --------(2)

(1) + (2) : 5a = 80

a = 16

when a = 16, b = 3

Example 3

Use the factor theorem to find the value of k for which (a + 2b) where a does not = 0 and b does not = 0, is a factor of a⁴ + 32b⁴ + a³b(k + 3).

Solution



Let f(a) = a⁴ + 32b⁴ + a³b(k + 3)

f(-2b) = (-2b)⁴ + 32b⁴ + (-2b)³b(k + 3) = 0

48b⁴ - 8b⁴(k+ 3) = 0

8b⁴[6 - (k + 3)] = 0

8b⁴(3 - k) = 0

Since b does not = 0, 3 - k = 0

k = 3

Example 4

Determine the value of k for which x + 2 is a factor of (x + 1)⁷ + (2x + k)³.

Solution



Let f(x) = (x + 1) + (2x + k)

f(-2) = (-2 + 1) + (-4 + k) = 0

(k - 4) - 1 = 0

(k - 4) = 1

k - 4 = 1

k = 5

Example 5

Given that f(x) = x³ - x² - ax - b, where a and b are constants, has the factor x - 3 but has a remainder of 13x - 11 when divided by x + 4. Calculate the values of a and b.

Hence (i) factorise f(x) completely. (ii) Solve the equation 27x³ - 3ax = ax² + b.

Solution



f(3) = 0

3³ - 3² - 3a - b = 0

3a + b = 18 --------(1)

f(-4) = (-4)³ - (-4)² - a(-4) - b = 13(-4) - 11

4a - b = 17 --------(2)

sub (1) into (2): a = 5, b = 3

(i) Factorising.

f(x) = (x - 3)(x² + 2x + 1)

= (x - 3)(x + 1)²

(ii) 27x³ - 9x² - 3ax - b = 0

(3x)³ - (3x)² - a(3x) - b = 0

Let y = 3x

y³ - y² - ay - b = 0

(y - 3)(y + 1)² = 0 [from (1)]

y = 3, y = -1

x = 1, x = -1/3

Example 6

Show that the expression x³ + (k-2) x² + (k-7)x - 4 has a factor x+1 for all values of k .If the expression also has a factor x+2, find the value of k and the third factor.



Solution

Let f(x) = x³ + (k-2) x² + (k-7)x - 4

f(-1) =(-1)³ + (k-2)(-1)² + (k-7)(-1) - 4 = -1+ k - 2 - k + 7 - 4 = 0

Therefore (x+1) is a factor of f(x).

x + 2 is a factor of f(x). -------(given)

Therefore f(-2)=0

(-2)³ + (k-2)(-2)² + (k-7)(-2) - 4 = 0

-8 + 4k -8 - 2k +14 - 4 = 0

2k = 6

k = 3

Therefore f(x) = x³ + x² - 4x - 4

Therefore (x + 1) (x + 2) = x² + 3x + 2 is also a factor of f(x).

Therefore the third factor of f(x) is x - 2.



Example 7

Given that kx³ + 2x² + 2x + 3 is a factor of kx³ - 2x + 9, have a common factor, what are the possible values of k?



Solution

Let f(x) = >kx³ + 2x² + 2x + 3 and g(x) = kx³ - 2x + 9.

Let (x - c) be a common factor of both f(x) and g(x).Therefore ...

f(c) = 0

kc³ + 2c² + 2c + 3 = 0 ---------(1)



g(c) = 0

kc³ - 2c + 9= 0 ---------(2)



Subtracting equation(2) from (1),we have

2c² + 4c - 6 = 0

c² + 2c - 3 = 0

(c + 3) (c - 1) =0

c = -3 or c = 1



Substituting c = -3 or c = 1 in equation (2), we get

k = 5/9 when c = -3 and

k = -7 when c = 1

The Factor Theorem

6 = 3 × 2

6÷2 => remainder = 0

6÷3 => remainder = 0

အၾကြင္း 0 ရေအာင္စားႏိုင္ေသာ စားကိန္းကို တည္ကိန္း၏ factor ဟုေခၚသည္။

The Factor Theorem:

Let f(x) be a polynomial. Then (x-k) is a factor of f(x) if and only if f(k) = 0.

Example 1

Find what value p must have in order that x - p may be a factor of
4x³ - (3p + 2)x² - (p² - 1)x + 3.

Let f(x) = 4x³ - (3p + 2)x² - (p² - 1)x + 3.
x - p is a factor of f(x) only if
f(p) = 0
4p³ - (3p + 2)p² - (p² - 1)p + 3 = 0
2p² - p - 3 = 0
(p + 1) (2p - 3) = 0
p = -1 or p = 3/2

Remainder Theorem ကိုအသံုးခ်တြက္ေသာ ပုစာၦမ်ား(3)

Example 10

The remainder when x⁴ + 3x² - 2x + 2 is divided by x + a is the square of the remainder when x² - 3 is divided by x + a, Calculate the possible values of a.
x⁴ + 3x² - 2x + 2 ကို x + a ႏွင့္စားေသာအခါ ရတဲ့ remainder သည္ x² - 3 ကို x + a ႏွင့္ စားေသာအခါ ရတဲ့ remainder၏ ႏွစ္ထပ္ႏွင့္ ညီလွ်င္ a ၏ တန္ဖိုးမ်ားကို ရွာပါ။

Let f(x) = x⁴ + 3x² - 2x + 2 and g(x) = x² - 3
By the problem,
f(-a) = [g(-a)]²
(-a)⁴ + 3(-a)² - 2(-a) + 2 =[(-a)² - 3]²
a⁴ + 3a² + 2a + 2 = a⁴ - 6a² + 9
9a² + 2a - 7 = 0
(a + 1) (9a - 7) = 0
a = -1 (or) a=7/9

Example 11

The expression ax³ - x² + bx - 1 leaves the remainder of -33 and 77 when divided by x + 2 and x - 3 respectively. Find the values of a and b and the remainder when divided by x - 2.
ax³ - x² + bx - 1 ကို x + 2 နဲ႔စားတဲ့အခါ -33 ရၿပီး x - 3 နဲ႔စားတဲ့အခါ 77 ရပါတယ္။ a ႏွင့္ b ကိုရွာပါ။ x - 3 ႏွင့္စားလို႔ရတဲ့ remainder ကိုလည္းရွာေပးပါ။

Let f(x) = ax³ - x² + bx - 1.
By the problem,
f(-2) = -33
a(-2)³ - (-2)² + b(-2) - 1 = -33
4a + b = -33 -------------(1)

f(3) = 77
a(3)³ - (3)² + b(3) - 1 = 77
9a + b = 29 -------------(2)

equation (2) - equation (1),we get
5a = 15 and a = 3

Substitute a = 3 in equation (1)
4(3) + b = 14
b = 2